A launching mechanism consists of an arm (uniform rod with mass 250. g and lengt

Question

A launching mechanism consists of an arm (uniform rod with mass 250. g and length 75.0 cm) with a holder (negligible mass) at one end mounted to a hub of radius 3.00 cm and negligible mass at the other end. A string is wrapped around the hub and passes to a spring of constant k = 2550 N/m that is unstretched when the launcher is in the vertical position. The launcher arm is pulled into the horizontal position and loaded with a 50.0 g point mass. It is released from rest and stops by hitting a pin at the vertical position. The projectile leaves the launcher with an initial horizontal velocity at a height of 2.00 m above the ground. Where does it land?

Explanation / Answer

When the rod is in horizontal postion , the spring is streched by 2piR/4 radias

so streching of the spring is = 2*3.14* 0.03.4 = 0.0471 m

Energy stored U = 0.5 kx^2

where K is sping contant

so

U = 0.5* 2550 * 0.0471* 0.0471 = 2.83 J

Moment of inertial of rod and mass I = mL^2 + ML^2/3

I = 0.05* 0.075* 0.75 + 025 * 0.75^2/3

I = 0.075 Kgm^2

tota energy of the system in vertical position = PE of the ball + PE of the rod

Etotal = (0.05 *9.81 * 0.75) + (0.25* 9.81 * 0.75/2)

= 1.28 J

now as total energy is conserved

2.83 - KE + PE

0.5 IW^2 = 2.83 -1.28 = 1.55 J

W = 6.43 rad/s

as V = r W

v = 0.75 * 6.43

v = 4.822 m/s

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